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kati45 [8]
2 years ago
5

In 2012, the population of a city was 6.29 million. The exponential growth rate was

Mathematics
1 answer:
RoseWind [281]2 years ago
4 0

9514 1404 393

Answer:

  a) P(t) = 6.29e^(0.0241t)

  b) P(6) ≈ 7.3 million

  c) 10 years

  d) 28.8 years

Step-by-step explanation:

a) You have written the equation.

  P(t) = 6.29·e^(0.0241·t)

__

b) 2018 is 6 years after 2012.

  P(6) = 6.29·e^(0.0241·6) ≈ 7.2686 ≈ 7.3 . . . million

__

c) We want t for ...

  8 = 6.29·e^(0.0241t)

  ln(8/6.29) = 0.0241t

  t = ln(8/6.29)/0.0241 ≈ 9.978 ≈ 10.0 . . . years

__

d) Along the same lines as the calculation in part (c), doubling time is ...

  t = ln(2)/0.0241 ≈ 28.7613 ≈ 28.8 . . . years

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judy has a piece of wood that is 4 5/8 feet long. she cuts off 3 feet 6 inches of the wood for a project. how much wood, in feet
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Read 2 more answers
Sharon purchased a used car for $24,600. The car depreciates exponentially by 8% per
Tju [1.3M]

Answer:

The value of car after 5 years of exponentially depreciation is $16,213.36  

Step-by-step explanation:

The initial amount of the car = i = $24,600

The depreciates rate of interest applied = r = 8%

Let The value of car after n years = A  

The number of years = n = 5 years

Now,According to question

The value of car after n years = initial price of car  × (1-\dfrac{\textrm rate}{100})^{\textrm time}

Or, The value of car after 5 years = i × (1-\dfrac{\textrm r}{100})^{\textrm n}

Or, $A = $24,600 × (1-\dfrac{\textrm 8}{100})^{\textrm 5}

Or, $A = $24,600 × (0.92)^{5}

Or, $A = $24,600 × 0.65908

Or, $A = $16,213.36

So, The value of car after 5 years = A = $16,213.36

Hence , The value of car after 5 years of exponentially depreciation is $16,213.36  . Answer

5 0
3 years ago
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